Qual Problem: When do Matrices Commute?

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The discussion focuses on proving that if A + B = AB for nxn matrices A and B, then AB = BA. The key approach involves manipulating the equation using the identity matrix, leading to the conclusion that (A-I)(B-I) = I, which implies B-I is the inverse of A-I. Participants also explore general conditions for matrix commutation, noting that two matrices commute if they are simultaneously diagonalizable, although this only applies if both matrices are diagonalizable. The conversation highlights the importance of algebraic manipulation and symmetry in deriving results about matrix properties. Overall, the thread provides insights into matrix relationships and conditions for commutation.
tornado28
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I'm preparing for a qualifying exam and this problem came up on a previous qual:

Let A and B be nxn matrices. Show that if A + B = AB then AB=BA.
 
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Hint: Consider (A-I)(B-I), where I is the nxn identity matrix.
 
Thanks morphism!

(A-I)(B-I) = AB-A-B+I = AB-AB+I=I. Therefore B-I is the inverse of A-I so we have that I=(B-I)(A-I) = BA-A-B+I = BA-AB+I. Thus BA-AB = 0 as needed.

How did you know to write it that way? Also, do you know any good general conditions related to matrices which commute? What is necessary for AB=BA? What (other than A+B=AB) is sufficient for AB=BA?
 
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I wrote it that way after fidgeting around with A+B=AB for a while. If you rewrite this as A+B-AB=0 then you might try to factor out A or B to get A(I-B)+B=0 or B(I-A)+A=0. The symmetry inspired me to subtract I from both sides of the first equation to get A(I-B)+B-I=-I <=> A(I-B)-(I-B)=-I <=> (I-A)(I-B)=I.

As for your other questions, I can't think of anything useful off the top of my head.
 
Thanks. I found another thread with information about when matrices commute. Apparently two matrices commute iff they're simultaneously diagonalizable.
 
tornado28 said:
Thanks. I found another thread with information about when matrices commute. Apparently two matrices commute iff they're simultaneously diagonalizable.
That's only true if the two matrices are diagonalizable to begin with! :)
 
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